C Looooops
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 18799 | Accepted: 4924 |
Description
A Compiler Mystery: We are given a C-language style for loop of type
I.e., a loop which starts by setting variable to value A and while variable is not equal to B, repeats statement followed by increasing the variable by C. We want to know how many times does the statement get executed for particular values of A, B and C, assuming that all arithmetics is calculated in a k-bit unsigned integer type (with values 0 <= x < 2k) modulo 2k.
for (variable = A; variable != B; variable += C)
statement;
I.e., a loop which starts by setting variable to value A and while variable is not equal to B, repeats statement followed by increasing the variable by C. We want to know how many times does the statement get executed for particular values of A, B and C, assuming that all arithmetics is calculated in a k-bit unsigned integer type (with values 0 <= x < 2k) modulo 2k.
Input
The input consists of several instances. Each instance is described by a single line with four integers A, B, C, k separated by a single space. The integer k (1 <= k <= 32) is the number of bits of the control variable of the loop and A, B, C (0 <= A, B, C < 2k) are the parameters of the loop.
The input is finished by a line containing four zeros.
The input is finished by a line containing four zeros.
Output
The output consists of several lines corresponding to the instances on the input. The i-th line contains either the number of executions of the statement in the i-th instance (a single integer number) or the word FOREVER if the loop does not terminate.
Sample Input
3 3 2 16 3 7 2 16 7 3 2 16 3 4 2 16 0 0 0 0
Sample Output
0 2 32766 FOREVER
Source
分析：
扩展欧几里得:
对于ax+by=c，求最小整数解
d=gcd(a,b)
1.先求ax0+by0=d的x0，y0。然后两边同乘d得a（x0*c/d）+b（y0*c/d）=gcd(a,b)*c/d=c，求出x=x0*c/d，y=y0*c/d
2.求最小解：由上求出特解x，最小解=(x%(b/d)+b/d)%(b/d)
参考学习网站：http://www.cnblogs.com/comeon4mydream/archive/2011/07/18/2109060.html
1 #include<cstdio> 2 #include<algorithm> 3 #include<iostream> 4 #include<string> 5 #include<cstring> 6 #include<vector> 7 using namespace std; 8 typedef long long ll; 9 ll exgcd(ll a,ll b,ll &x,ll &y){//扩展欧几里得算法 10 if(b==0){ 11 x=1; 12 y=0; 13 return a; 14 } 15 ll d=exgcd(b,a%b,x,y); 16 ll t=y; 17 y=x-a/b*y; 18 x=t; 19 return d; 20 } 21 int main(){ 22 ll A, B, C, k; 23 while(cin>>A>>B>>C>>k&&(A||B||C||k)){ 24 ll b=1ll<<k,x,y; 25 ll c=B-A,a=C; 26 ll d=exgcd(a,b,x,y); 27 if(c%d){ 28 cout<<"FOREVER"<<endl; 29 } 30 else{ 31 b=b/d;//如果少了这歩，最终求得的就可能不是最小整数解，而是某一特解。可以仔细想一想为什么b要除以d 32 //cout<<d<<endl; 33 cout<<((x*c/d%b+b)%b)<<endl;//求最小整数解 34 } 35 } 36 return 0; 37 }